question_answer 10)

The threshold frequency \[{{v}_{0}}\] for a metal is\[7\times {{10}^{14}}{{\sec }^{-1}}\]. Calculate the kinetic energy of an electron emitted when radiation of frequency \[v=1\times {{10}^{15}}{{\sec }^{-1}}\] hits the metal.

Answer:

Information shadow: Threshold frequency, \[{{v}_{0}}=7\times {{10}^{14}}{{\sec }^{-1}}\] Frequency of incident radiation, \[v=1\times {{10}^{15}}{{\sec }^{-1}}\] Problem solving strategy: Absorbed = Threshold + Kinetic energy energy energy of photoelectron \[hv~~~~~~~=~~~~~~~~~~~~h{{v}_{0}}~~~~~~~~~~+~~~~~~~~~~~~~~~~~~~~K.E.\] Working it out: \[K.E.=hv-h{{v}_{0}}=h(v-{{v}_{0}})\] \[=6.626\times {{10}^{-34}}(1\times {{10}^{15}}-7\times {{10}^{14}})\] \[=6.626\times {{10}^{-34}}\times 3\times {{10}^{14}}\] \[=1.988\times {{10}^{-19}}J\]